0
$\begingroup$

Using pythagoras theorem, I received.. $$l (Slant)=\sqrt{r^2+h^2}$$ Using the volume of a cone formula in terms of $h$..$$h=\dfrac{54}{r^2}$$ I then subbed this into the 1st equation and diffrenciated with respect to $r$..$$\dfrac{\mathrm dl}{\mathrm dr}=\dfrac{2r-\dfrac{11664}{r^5}}{2\sqrt{(r^2+\dfrac{2916}{r^4})}}$$

Looking at it now I probably should have replaced 54 with a symbol, can someone tell me if I'm on the right track? I put it equal to zero then transposed in terms of $r$ and received $1.45$ for the radius. I subbed that value back into the $l$ equation on top and recieved the wrong answer...the answer is meant to be $3\sqrt{3}$

If there's a quicker way Please advise with some hints.

$\endgroup$
1
$\begingroup$

You had the result in your hands.

$2r-\dfrac{4*54^2}{r^5}=0$ means $r^6=2*54^2=2^3*(3^2)^3$, $r^2=2*3^2=18$

Hence $h^2=9$ and $l^2=18+9$, $l=3\sqrt 3$.

$\endgroup$
2
  • $\begingroup$ I made a silly error, typed into my calculator incorrectly..thanks for the answer! $\endgroup$
    – Modrisco
    Jun 17 '15 at 11:24
  • $\begingroup$ @Modrisco You should avoid if possible to use calculator! Usually problems are made to get "nice" results, and this means simplifications to look for around. $\endgroup$
    – Martigan
    Jun 17 '15 at 11:55
0
$\begingroup$

$$l^2=r^2+\left(\dfrac{54}{r^2}\right)^2=\dfrac{r^2}2+\dfrac{r^2}2+\left(\dfrac{54}{r^2}\right)^2$$

Now using AM-GM inequality for three variables,

$$\dfrac{\dfrac{r^2}2+\dfrac{r^2}2+\left(\dfrac{54}{r^2}\right)^2}3\ge\sqrt[3]{\dfrac{r^2}2\cdot\dfrac{r^2}2\cdot\left(\dfrac{54}{r^2}\right)^2}$$

The equality occurs if $\dfrac{r^2}2=\dfrac{r^2}2=\left(\dfrac{54}{r^2}\right)^2$

As $r\ge0,$ we need $\dfrac r{\sqrt2}=\dfrac{54}{r^2}\iff r=3\sqrt2$

$\endgroup$
3
  • $\begingroup$ $$\implies\dfrac{l^2}3\ge \sqrt[3]{27^2}=(3^6)^{1/3}=9$$ $\endgroup$ Jun 17 '15 at 11:08
  • $\begingroup$ AM-GM inequality for 3 variables? I'm not sure what's going on here...is there another more simpler way? I'm not that advanced in mathematics. $\endgroup$
    – Modrisco
    Jun 17 '15 at 11:18
  • $\begingroup$ @Modrisco, See en.wikipedia.org/wiki/… This helps you avoiding calculus $\endgroup$ Jun 17 '15 at 11:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.